**Theorem 12 (Gromov)**: Let be torsion-free -hyperbolic group. If such that , then for all sufficiently large , .

**Remark:** The torsion-free hypothesis is not necessary, but it allows us to avoid some technicalities. For instance, it is a non-obvious fact that an infinite hyperbolic group contains a copy of .

For the rest of this lecture will be a torsion-free -hyperbolic group, where are primitive (i.e. not proper powers).

Recall that for torsion-free -hyperbolic, primitive implies that .

If and do not commute we can show there is some point on arbitrarily far from .

Hence we have the following lemma.

**Lemma 13: **

If and do not commute there is some point on arbitrarily far from .

**Proof:** Suppose not. That means such that such that . So is in . But the Cayley graph is locally finite so has finitely many elements. By the Pigeonhole Principle such that for some . Then . But then . .

For a moment view and as the horizontal and vertical geodesics in . For two points on and on , we can argue that the geodesic between them curves toward the origin.

And so we have Lemma 14.

**Lemma 14:** There exists such that , .

**Proof**:

Recall that by is a quasi-isometric embedding. So by Theorem 6, and

By Lemma 13 choose such that

. Choose such that . Now, must be -close to so for some point on the geodesic between and , . Then .

For a subgroup , one can choose a closest point projection which is -equivariant. (Write . Choose where and are close and declare to be -equivariant.) is typically not a group homomorphism.

We’re interested in and .

In , there is some such that either or .

**Lemma 15:** such that , or .

**Proof:**

Let . WLOG, is -close to and since is the closest point to (in particular compared to ). So . .

Now we can prove the theorem.

**Proof of Theorem 12:**

The idea is to use the Ping-Pong Lemma on the Cayley graph.

Let and let , where is provided by Lemma 15. For all we have and likewise for all we have . In particular, .

Let . By -equivariance,

for any . In particular,

by the triangle inequality. Similarly,

for all and all . Because and are quasi-isometrically embedded, it follows that and for .

Therefore, by the Ping-Pong Lemma .

## 2 comments

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4 March 2009 at 6.49 am

Henry WiltonWow! That’s a lot of mathematics and a lot of pictures. Many many thanks, to both Sam and Emily. Great job. I tidied up the proof of Theorem 12 a little.

2 May 2011 at 10.55 am

The Tits alternative and non-positive curvature | Here there be dragons[…] elementary proof of the above in the case where the subgroup is torsion free is presented on Henry Wilton’s geometric group theory blog (a precursor and inspiration for the blog […]